Quiz**
Determine whether f is a bijection from Z to Z if f(x) = $x^{5}$ + 1? Also determine if it is invertible or not.
I gave an answer of " f is not a bijection and is invertible" It was wrong. Please explain
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Consider $3 \in \mathbb{Z}$. If $f$ were onto we would be able to find some $n \in \mathbb{Z}$ such that $f(n)=n^5+1=3$. But that implies $n=\sqrt[5]{2} \notin \mathbb{Z}$. The point of this was to show there is an element in the range of $f$ that does not get mapped to by any element in the domain, meaning $f$ is not onto.
As you already stated, we now know $f$ is not a bijection. Further, being a bijection is a necessary condition of being invertible, so we may also conclude that $f$ is not invertible.
graydad
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Let $X, Y$ be nonempty subsets of $\mathbb{Z}$. A function $f: X \to Y$ is injective if $f(x_{1}) = f(x_{2})$ implies $x_{1} = x_{2}$; surjective if $f(X) = Y;$ and bijective if $f$ is both injective and surjective.
Since the domain of the function $$x \mapsto x^{5} + 1$$ is $\mathbb{Z}$, and since if $x_{1}^{5} + 1 = x_{2}^{5} + 1$ then $x_{1} = x_{2}$, the function $x \mapsto x^{5} + 1$ is injective.
Nevertheless, if $x^{5} + 1 = 4$ then $x$ is not an integer, whence the range of $x \mapsto x^{5} + 1$ is a proper subset of $\mathbb{Z}$. Thus, $x \mapsto x^{5}+1$ is not surjective, so that it is not bijective, and hence it is not invertible.
Yes
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